已知2x+y-2≥0,x-2y+4≥0,3x-y-3≤0,求z=|2x+y+5|的最大值和最小值

由2x+y-2≥0,x-2y+4≥0得3x-y+2≥0,即3x-y≥-2,又由3x-y-3≤0得3≥3x-y≥-2；x-2y+4≥0得-x+2y≤4,所以 z最大=|2x+y+5|=|3x-y-x+2y+5|=|(3x-y)+(-x+2y)+5|≤|3+4+5|=12,Z最小=|2x+y+5|=|3x-y-x+2y+5|=|(3x-y)-(x-2y)...“Z最小=|2x+y+5|=|3x-y-x+2y+5|=|(3x-y)-(x-2y)+5|≤|-2-4+5|=1”这一点是不是出错了？前面有-x+2y≤4，所以可以知道x-2y≥-4也就是说|(3x-y)-(x-2y)+5|≤|-2-（-4）+5|应该等于7吧是。应该为Z最小=|2x+y+5|=|（2x+y-2）+7|≥||0+7|=7